The downvote was probably because your question was not of research level. You asked for clarifying a certain step in a certain proof of a well-known fact. E.g. see my response below for two more proofs available in introductory textbooks.
–
GH from MOFeb 20 '12 at 19:44

2 Answers
2

Since you edited the question but did not say that it is clear now, I assume you are hoping for some details in addition to what Ralph said. So:

Let $f_n(s) = \frac{1}{n^s}- \frac{1}{(n+1)^{s}}$ the derivative of this with respect to $s$ is as you computed $\frac{-\log n}{n^s}+ \frac{\log(n+1)}{(n+1)^{s}}$. This is a function of $s$, and you do not consider it on $[n,n+1]$, but rather $[1,s]$ for each $n$.

The MVT tells you that $\frac{f_{n}(s) - f_n(1)}{s -1} = f'(s_n) $ for some $s_n\in [1,s]$.
Multiplying by $(s-1)$ and plugging in the explicit expressions for the functions this means
$$(\frac{1}{n^s}- \frac{1}{(n+1)^{s}}) - (\frac{1}{n} -\frac{1}{n+1}) = (s-1)(\frac{-\log n}{n^{s_n}}+ \frac{\log (n+1)}{(n+1)^{s_n}}).$$
The left hand side appears in the original sum, and the result is obtained by instead plugging in the right hand side.

It is a standard fact that $L(s,\chi)$, initially defined as a locally uniformly convergent Dirichlet series in $\Re s>1$, extends to a holomorphic function on $\mathbb{C}$. See for example Chapter 9 in Davenport: Multiplicative Number Theory, especially the first half of Page 69.

Now basic complex analysis tells us that the Taylor series of $L(s,\chi)$ around $s=1$ converges absolutely on $\mathbb{C}$, of which
$$ L(s,\chi)=L(1,\chi)+O(|s-1|) $$
is a consequence. In fact for the last equation we only need to know that $L'(s,\chi)$ is continuous on $\mathbb{C}$, which follows directly from Cauchy's integral formula.

A quick proof of the holomorphicity of $L(s,\chi)$ in $\Re s>0$ follows from the fact that the partial sums $\sum_{n\leq x}\chi(n)$ are bounded. See Proposition 9 in Section VI.2 of Serre: A course in arithmetic, or Theorem 1.3 in Montgomery-Vaughan: Multiplicative number theory I.